Question-1 :2022-nov-Chemistry_paper_1__TZ0_SL
Topic:
Discuss: How many oxygen atoms are present in $0.0500 \mathrm{~mol} ~\mathrm{Ba}(\mathrm{OH})_2 \cdot 8 \mathrm{H}_2 \mathrm{O}$ ?
$$
N_{\mathrm{A}}=6.02 \times 10^{23}
$$
A. $3.01 \times 10^{23}$
B. $6.02 \times 10^{23}$
C. $3.01 \times 10^{24}$
D. $6.02 \times 10^{24}$
Answer/Explanation
Solution:
The compound $\mathrm{Ba}(\mathrm{OH})_2 \cdot 8 \mathrm{H}_2 \mathrm{O}$ contains 10 oxygen atoms, two from $\mathrm{OH}$ group, and eight oxygen atoms in each water molecule, for a total of $2+8=10$ oxygen atoms.
To find the number of oxygen atoms in $0.0500 \mathrm{~mol}$ of the compound, we can use Avogadro’s number ($N_{\mathrm{A}}$) to convert from moles to number of atoms, and then multiply by the number of oxygen atoms present in formula.
The calculation is as follows:
\begin{align*}
\text{Total no. of atoms} &= \frac{0.0500 \mathrm{~mol}}{1} \times \frac{N_{\mathrm{A}}}{1 \mathrm{~mol}}\\
&= 0.0500 \times N_{\mathrm{A}}
\end{align*}
Then, the total number of oxygen atoms in $0.0500 \mathrm{~mol}$ of the compound is:
\begin{align*}
\text{Number of oxygen atoms} &= \text{Total no. of atoms } \times \text{Number of oxygen atoms}\\
&= 0.0500 \times N_{\mathrm{A}} \times 10\\
&= 3.01 \times 10^{23}
\end{align*}
$\colorbox{yellow}{Correct Option: (A) $3.01 \times 10^{23}$ oxygen atoms.}$
Question-2 :2022-nov-Chemistry_paper_1__TZ0_SL
Topic:
Discuss: What is the change of state for a gas to a solid?
A. Condensation
B. Deposition
C. Freezing
D. Sublimation
Answer/Explanation
Solution:
The change of state for a gas to a solid is called “Deposition”. During deposition, the gas molecules lose energy and come together to form a solid without passing through the liquid phase. This is the opposite process of sublimation, which is the change of state from a solid directly to a gas. Condensation is the change of state from a gas to a liquid, while freezing is the change of state from a liquid to a solid.
Question-3 :2022-nov-Chemistry_paper_1__TZ0_SL
Topic:
Discuss: How many moles of carbon dioxide are produced by the complete combustion of $7.0 \mathrm{~g}$ of ethene, $\mathrm{C}_2 \mathrm{H}_4(\mathrm{~g})$ ?
$$
M_{\mathrm{r}}=28
$$
A. $0.25$
B. $0.5$
C. $0.75$
D. $1.0$
Answer/Explanation
Solution:
The balanced equation for the combustion of ethene is:
$\mathrm{C}_2 \mathrm{H}_4+3 \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}$
From the balanced equation, we see that 2 moles of $\mathrm{CO}_2$ are produced for every 1 mole of ethene consumed.
First, we need to calculate the number of moles of ethene in 7.0 g:
$n\left(\mathrm{C}_2 \mathrm{H}_4\right)=\frac{m}{M_{\mathrm{r}}}=\frac{7.0 \mathrm{~g}}{28 \mathrm{~g} / \mathrm{mol}}=0.25 \mathrm{~mol}$
Since 2 moles of $\mathrm{CO}_2$ are produced for every 1 mole of ethene consumed, the number of moles of $\mathrm{CO}_2$ produced will be:
$n\left(\mathrm{CO}_2\right)=2 \times n\left(\mathrm{C}_2 \mathrm{H}_4\right)=2 \times 0.25 \mathrm{~mol}=0.5 \mathrm{~mol}$
$\colorbox{yellow}{Correct Option : B}$
Question-4 :2022-nov-Chemistry_paper_1__TZ0_SL
Topic:
Discuss: Which is a possible empirical formula for a substance with $M_{\mathrm{r}}=42 ?$
A. $\mathrm{CH}$
B. $\mathrm{CH}_2$
C. $\mathrm{C}_3 \mathrm{H}_6$
D. $\mathrm{C}_3 \mathrm{H}_8$
Answer/Explanation
Solution:
To find the empirical formula of a substance, we need to determine the smallest whole-number ratio of atoms in the compound. Here’s how we can do it:
- Find the molar mass of the compound, $M_{\mathrm{r}}.$
- Assume a simple empirical formula for the compound, such as $\mathrm{C}_x\mathrm{H}_y,$ where $x$ and $y$ are unknown integers.
- Calculate the molar mass of the empirical formula by multiplying the atomic mass of each element by its subscript and adding the results: $M_{\mathrm{r}}(\mathrm{C}_x\mathrm{H}_y) = x \cdot M_{\mathrm{r}}(\mathrm{C}) + y \cdot M_{\mathrm{r}}(\mathrm{H}).$
- Solve for $x$ and $y$ by setting $M_{\mathrm{r}}(\mathrm{C}_x\mathrm{H}y) = M_{\mathrm{r}}$ and finding the smallest whole-number ratio of $x$ and $y$ that satisfies the equation.
In this problem, we’re given $M_{\mathrm{r}} = 42,$ and we assume a simple empirical formula of $\mathrm{C}_x\mathrm{H}_y.$ We can use the atomic masses of carbon and hydrogen to calculate the molar mass of the empirical formula:
$M_{\mathrm{r}}(\mathrm{C}_x\mathrm{H}_y) = x \cdot M{\mathrm{r}}(\mathrm{C}) + y \cdot M{\mathrm{r}}(\mathrm{H}) = x \cdot 12\mathrm{\ g/mol} + y \cdot 1\mathrm{\ g/mol}.$
Setting $M_{\mathrm{r}}(\mathrm{C}_x\mathrm{H}_y) = 42$ and solving for $x$ and $y,$ To find the smallest whole-number ratio of $x$ and $y,$ we can try different values of $y$ until we get an integer value of $x.$
For $y = 6,$ we get: $x = 3$
$\mathrm{C}_3\mathrm{H}_6= M_{\mathrm{r}}$
Molecular Formula $=($ Empirical Formula $)_ \mathrm{x}$
$\colorbox{yellow}{$\therefore$ Empirical Formula $=\left(\mathrm{CH}_2\right)$ (B) }$
Question-5 :2022-nov-Chemistry_paper_1__TZ0_SL
Topic:
Discuss: Which quantities are different between two species represented by the notation ${ }_{52}^{128} \mathrm{Te}$ and ${ }_{53}^{128} \mathrm{I}^{-}$?
A. The number of protons only
B. The number of protons and electrons only
C. The number of protons and neutrons only
D. The number of protons, neutrons and electrons
Answer/Explanation
Solution:
The notation ${ }{52}^{128} \mathrm{Te}$ represents the isotope of tellurium with 52 protons, 76 neutrons, and 52 electrons, while the notation ${ }{53}^{128} \mathrm{I}^{-}$ represents the iodide ion, which has 53 protons, 75 neutrons, and 54 electrons.
Therefore, the quantities that are different between these two species are:
- The number of protons (52 vs 53)
- The number of neutrons (76 vs 75)
- The number of electrons (52 vs 54)
The number of protons, neutrons and electrons are different between the two species.
$\colorbox{yellow}{Correct Option : D}$